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KingGeorge Group Title

[SOLVED] Here's a cute predicate logic problem. Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    is xyzw a combination of digit or multiplication of digit??

    • 2 years ago
  2. KingGeorge Group Title
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    multiplication

    • 2 years ago
  3. experimentX Group Title
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    lol .. didn't realize they were complex

    • 2 years ago
  4. KingGeorge Group Title
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    It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

    • 2 years ago
  5. KingGeorge Group Title
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    Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.

    • 2 years ago
  6. KingGeorge Group Title
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    Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]

    • 2 years ago
  7. experimentX Group Title
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    \[ \color{white}{\text{LOL i didn't know that!!.}} \]

    • 2 years ago
  8. KingGeorge Group Title
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    Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]

    • 2 years ago
  9. nbouscal Group Title
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    Choose y=0, then choose w=-(x+z).

    • 2 years ago
  10. KingGeorge Group Title
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    Bingo.

    • 2 years ago
  11. experimentX Group Title
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    But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

    • 2 years ago
  12. KingGeorge Group Title
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    \(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

    • 2 years ago
  13. experimentX Group Title
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    then it must be false ... i guess so!!

    • 2 years ago
  14. KingGeorge Group Title
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    The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.

    • 2 years ago
  15. experimentX Group Title
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    does't y=0 make statement untrue?? xyzw=0, for y=0

    • 2 years ago
  16. nbouscal Group Title
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    Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).

    • 2 years ago
  17. KingGeorge Group Title
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    If \(y\) is 0, and \(x=-(x+z)\) we have \[x+0+z+(-(x+z))=x+z-x-z=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]

    • 2 years ago
  18. experimentX Group Title
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    Oh ... stupid me ... i never considered the negative values!!

    • 2 years ago
  19. KingGeorge Group Title
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    Hope everyone enjoyed this little (unintuitive) problem. Good night.

    • 2 years ago
  20. experimentX Group Title
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    thanks!! i'll try to think broader way next time!!

    • 2 years ago
  21. KingGeorge Group Title
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    I'll try to get a problem just for you next time then. Any preference about the subject?

    • 2 years ago
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